# Water is being boiled in an open kettle that has a 0.500-cm-thick circular aluminum bottle with a radius of 12.0cm. If the water boils away at a rate

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Water is being boiled in an open kettle that has a 0.500-cm-thick circular aluminum bottle with a radius of 12.0cm. If the water boils away at a rate of 0.500 kg/min,what is the temperature of the lower surface of the bottom of the kettle? Assume that the top surface of the bottom of the kettle is at $$\displaystyle{100}^{\circ}$$ C.

2021-03-29
Water is being boiled in an open kettlethat has a (L = 0.500-cm) -thick circular aluminum bottle with a radius of (r = 12.0cm).
If the water boils away at a rate of
$$\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}={0.500}\ {k}\frac{{g}}{\min}={8.33}\cdot{10}^{{-{3}}}\ {k}\frac{{g}}{{s}}$$
Let the temperature of the lower surface of the bottom of thekettle be $$\displaystyle{{t}_{{2}}^{\circ}}$$ C
The temperature of thetop surface of the bottom of the kettle ($$\displaystyle{t}_{{1}}={100}^{\circ}{C}$$)
Since the heat given by the aluminium bottle = heattaken by boiled water
$$\displaystyle{k}{A}{\left({t}_{{2}}-{t}_{{1}}\right)}\cdot{\frac{{{\left.{d}{t}\right.}}}{{{L}}}}{)}={d}{m}\cdot{L}_{{s}}$$
Where
$$\displaystyle{K}={238}\ \frac{{J}}{{s}}$$
$$\displaystyle{A}={p}{r}^{{2}}$$
dt=time taken for the mass dm
Latent Heat of steam ($$\displaystyle{L}_{{s}}$$)=2256000 J/kg
Ahain from above equation
$$\displaystyle{k}{A}{\frac{{{t}_{{2}}-{t}_{{1}}}}{{{d}}}}={\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}\cdot{L}_{{s}}$$
Now calculate the value of $$\displaystyle{t}_{{2}}$$ from above